Chem 12 Practice Worksheet - Answer Key

Chem 12

Practice Worksheet - Answer Key

Redox #1 (KEY)

1. Explain the meaning of each of the following terms:

a) oxidation

a half-reaction that involves the loss of electron(s)

b) reduction

a half-reaction that involves the gain of electron(s)

c) reducing agent

a species that causes another to be reduced; it itself is oxidized

d) oxidizing agent

a species that causes another to be oxidized; it itself is reduced

2. Identify each of the following as examples of oxidation of reduction:

a) Li(s)

Li+(aq) + e?

= oxidation

?

2?

b) S(s) + 2e

S (aq)

= reduction

3+

?

2+

c) Cr (aq) + e

Cr (aq)

= reduction

2+

4+

?

d) Sn (aq)

Sn (aq) + 2e

= oxidation

3. Given the following redox reaction:

a)

b)

c)

d)

Fe(s) + 2 Ag+(aq)

Which species is the reducing agent?

Which species is the oxidizing agent?

Which species is the being oxidized?

Which species is the being reduced?

4. Given the following redox reaction:

Fe2+(aq) + 2 Ag(s)

Fe

Ag+

Fe

Ag+

3 Zn2+(aq) + 2 Al(s)

a) Identify, and write the halfreaction for oxidation and the halfreaction for reduction.

oxidation = Al(s)

b) Name the reducing agent and the

oxidizing agent for the reaction

RA = Al

3 Zn(s) + 2 Al3+(aq)

Al3+(aq) + 3e?

reduction = Zn2+(aq) + 2e?

Zn(s)

OA = Zn2+

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5. Identify the material being reduced, the material being oxidized, the reducing agent, and

the oxidizing agent in each of the following redox reactions. Try to write the half-cell

reactions.

a) 2 Na(s) + ? O2(g)

Na2O(s)

b)

c)

reduction = ? O2(g) + 2e?

O2?

oxidation = Na(s)

Na+(aq) + e?

Fe(s) + Cu2+(aq)

Fe2+(aq) + Cu(s)

O2 = reduced; OA

Na = oxidized; RA

reduction = Cu2+(aq) + 2e?

Cu(s)

2+

Fe (aq) + 2e?

oxidation = Fe(s)

Cu2+ = reduced; OA

Fe = oxidized; RA

Sn4+(aq) + Fe2+(aq)

Sn2+(aq) + Fe3+(aq)

reduction = Sn4+(aq) + 2e?

Sn2+(aq)

oxidation = Fe2+(aq)

Fe3+(aq) + e?

d)

2 Mg(s) + O2(g)

2 MgO(s)

reduction = O2 + 4e?

2 O2?

oxidation = Mg(s)

Mg2+ + 2e?

e)

Ca(s) + Cl2(g)

2 H2(g) + O2(g)

O2 = reduced; OA

Mg = oxidized; RA

CaCl2(s)

reduction = Cl2(g) + 2e?

2 Cl?

oxidation = Ca(s)

Ca2+(aq) + 2e?

f)

Sn4+ = reduced; OA

Fe2+ = oxidized; RA

Cl2 = reduced; OA

Ca = oxidized; RA

2 H2O(l)

reduction = O2(g) + 4e?

2 O2?

oxidation = H2(g)

2 H +(aq) + 2e?

O2 = reduced; OA

H2 = oxidized; RA

6. When copper wire is placed in a solution of silver nitrate, what colour does the solution turn

after some time? Without looking at your notes, write the balanced half-reactions for the

oxidation and reduction processes that are occurring.

The solution turns blue, because Cu2+ is being formed.

Reduction = Ag+(aq)

Ag(s) + e?

Oxidation = Cu(s)

Cu2+(aq) + 2e?

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Chem 12

Redox #2 (KEY)

1. Determine the oxidation number of all atoms in each of the following compounds and

polyatomic ions.

a)

SO2

b)

CaF2

c)

PO33?

d)

PO43?

e)

H2O2

f)

NaSiO4

g)

KMnO4

S = +4

O = ¨C2

Ca = +2

F = ¨C1

h)

MnO2?

i)

KHSO3

P = +3

O = ¨C2

P = +5

O = ¨C2

j)

S2O72?

k)

NiSO4

H = +1

O = ¨C1

**peroxide

Na = +1

Si = +7

O = ¨C2

K = +1

Mn = +7

O = ¨C2

l)

SO3

m)

n)

Mn = +3

O = ¨C2

K = +1

H = +1

S = +4

O = ¨C2

S = +6

O = ¨C2

Ni = +2

S = +6

O = ¨C2

S = +6

O = ¨C2

o)

N2O3

p)

NO+

q)

NH3

r)

N2O

s)

H2CrO4

S2O7

S = +7

O = ¨C2

t)

P2O5

H2S2O3

H = +1

S = +2

O = ¨C2

u)

O3

N = +3

O = ¨C2

N = +3

O = ¨C2

N = ¨C3

H = +1

N = +1

O = ¨C2

H = +1

Cr = +6

O = ¨C2

P = +5

O = ¨C2

O = zero

2. Identify the species oxidized and reduced, as well as the oxidizing agent and the reducing

agent for the following reactions:

a)

CH4 + 2 O2

CO2 + 2 H2O

oxidized/RA = CH4 (Ox # of C goes from ¨C4 to +4)

reduced/OA = O2 (Ox # of O goes from 0 to ¨C2)

b)

Cr2O72? + 2 OH?

2 CrO42? + H2O

not a redox (oxidation number of Cr stays at +6)

c)

O3 + NO

O2 + NO2

oxidized/RA = NO (Ox # of N goes from +2 to +4)

reduced/OA = O3 (Ox # of O goes from 0 to ¨C2)

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d)

2 H2O2

2 H2O + O2

oxidized/RA = O in H2O2 (Ox # of O goes from ¨C1 to 0 in O2)

reduced/OA = O in H2O2 (Ox # of O goes from ¨C1 to ¨C2 in H2O)

e)

2 CuCl

CuCl2 + Cu

oxidized/RA = Cu in CuCl (Ox # of Cu goes from +1 to +2)

reduced/OA = Cu in CuCl (Ox # of Cu goes from +1 to 0)

f)

HCl + NH3

NH4Cl

not a redox (oxidation numbers did not change for all atoms)

g)

SiCl4 + 2 H2O

4 HCl + SiO2

not a redox (oxidation number of Si stays at +4)

h)

I? + ClO?

I3? + Cl?

oxidized/RA = I? (Ox # of I goes from ¨C1 to ? )

reduced/OA = ClO? (Ox # of Cl goes from +1 to ¨C1)

i)

As2O3 + MnO3?

H3AsO4 + Mn2+

oxidized/RA = As2O3 (Ox # of As goes from +3 to +5)

reduced/OA = MnO3? (Ox # of Mn goes from +5 to +2)

j)

Mn2+ + NaBiO3

Bi3+ + MnO4?

oxidized/RA = Mn2+ (Ox # goes from +2 to +7)

reduced/OA = NaBiO3 (Ox # of Bi goes from +5 to +3)

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Chem 12

Redox #3 (KEY)

1. Balance the following half-reactions

ACIDIC

a) Ce4+ ? Ce2+

b) I2 ? I¨C

c) I2 ? I3¨C

d) Mn2+ ? MnO2

e) O2 ? H2O2

f) S2O82¨C ? HSO4¨C

g) H3AsO4 ? HAsO2

h) H2SeO3 ? Se

i) CH3COOH ? C2H5OH

j) I¨C + ClO¨C ? I3¨C + Cl¨C

k) Br¨C + MnO4¨C ? Br2 + Mn2+

*** reaction (j) is a full redox.

BASIC

l) N2H4 ? N2

m) HO2¨C ? O2

n) HXeO4¨C ? HXeO63¨C

o) Cr(OH)3 ? CrO42¨C

p) CH3CHO ? CH2CH2

q) Al + MnO4¨C ? MnO2 + Al(OH)4¨C

r) Cl2 ? Cl¨C + OCl¨C

s) NO2¨C + Al ? NH3 + AlO2¨C

*** reactions (q) & (r) are full redox.

Ce4+ + 2e¨C ? Ce2+

I2 + 2e¨C ? 2 I¨C

3 I2 + 2e¨C ? 2 I3¨C

Mn2+ + 2 H2O ? MnO2 + 4 H+ + 2e¨C

O2 + 2 H+ + 2e¨C ? H2O2

S2O82¨C + 2 H+ + 2e¨C ? 2 HSO4¨C

H3AsO4 + 2 H+ + 2e¨C ? HAsO2 + 2 H2O

H2SeO3 + 4 H+ + 4e¨C ? Se + 3 H2O

CH3COOH + 4 H+ + 4e¨C ? C2H5OH + H2O

3 I¨C + ClO¨C + 2 H+ ? I3¨C + Cl¨C + H2O

2 Br¨C + MnO4¨C + 8 H+ + 3e¨C ? Br2 + Mn2+ + 4 H2O

N2H4 + 4 OH¨C ? N2 + 4 H2O + 4e¨C

HO2¨C + OH¨C ? O2 + H2O + 2e¨C

HXeO4¨C + 4 OH¨C ? HXeO63¨C + 2 H2O + 2e¨C

Cr(OH)3 + 5 OH¨C ? CrO42¨C + 4 H2O + 3e¨C

CH3CHO + H2O + 2e¨C ? CH2CH2 + 2 OH¨C

Al + MnO4¨C + 2 H2O ? MnO2 + Al(OH)4¨C

2 OH¨C + Cl2 ? Cl¨C + OCl¨C + H2O

NO2¨C + Al + 3 H2O + 3e¨C ? NH3 + AlO2¨C + 3 OH¨C

2. Balance the following and calculate the change in oxidation number for the species oxidized

or reduced. Compare this to the number of electrons in the balanced reaction.

a)

NO2¨C

NO3¨C

+3

+5

loss of 2e¨C

balanced: NO2¨C + H2O

NO3¨C + 2 H+ + 2e¨C

b)

MnO4¨C

MnO2

+7

+4

gain of 3e¨C

balanced: MnO4¨C + 4 H+ + 3e¨C

MnO2 + 2 H2O

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